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2 days ago

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Charge is distributed uniformly along a long straight wire. The electric field 2.00 cm from the wire is 20.0 NC/ directed radial

ly inward towards the axis of symmetry. The linear charge density on the wire is

1 answer:

Maru [2.9K]2 days ago

7 0

The linear charge density along the wire is 4.45 × 10^-11 C/m.

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A charge of uniform volume density (40 nC/m3) fills a cube with 8.0-cm edges. What is the total electric flux through the surfac

Keith_Richards [2877]

Answer:

The flux across the cube's surface is $2.314\ Nm^{2}/C$ .

Solution:

According to the details provided:

Cube edge length, a = 8.0 cm = $8.0\times 10^{- 2}\ m$ .

Volume charge density, $\rho_{v} = 40 nC/m^{3} = 40\times {- 9}\ C/m^{3}$ .

Now,

To find the electric flux:

$\phi = \frac{q}{\epsilon_{o}}$

where

$\phi$ = electric flux

$\epsilon_{o} = 8.85\times 10^{- 12}\ F/m$ = permittivity of vacuum.

The volume charge density for this scenario is described by:

$\rho_{v} = \frac{Total\ charge, q}{Volume of cube, V}$

Cube volume, $V = a^{3}$ .

Thus,

$V = (8.0\times 10^{- 2})^{3} = 5.12\times 10^{- 4}\ m^{3}$ .

The total charge can be derived from equation (2):

$q = \rho_{v}V = 40\times {- 9}\times 5.12\times 10^{- 4}$ .

$q = 2.048\times 10^{-11}\ F = 20.48\ pF$ .

Now, insert the value of 'q' into equation (1):

$\phi = \frac{2.048\times 10^{-11}}{8.85\times 10^{- 12}} = 2.314\ Nm^{2}/C$ .

5 0

1 month ago

A good quarterback can throw a football at 27 m/s (about 60 mph). If we assume that the ball is caught at the same height from w

ValentinkaMS [3076]

Response:

The ball remained airborne for 3.896 seconds

Explanation:

Given that

g = 9.8 m/s², representing gravitational acceleration,

If the angle of launch is 45°, the horizontal range will be maximized.

Both horizontal and vertical launch velocities are equal, each equating to

v_h = v cos θ

v_h = 27 × cos 45°

= 19.09 m/s.

The duration to reach maximum height is half of the flight time.

v = u + at ∵ v = 0 (at maximum height)

19.09 - 9.8 t₁ = 0

t₁ = 1.948 s

The total time in the air equals twice the time to reach maximum height

2 t₁ = 3.896 s

The horizontal distance covered is

D = v × t

D = 3.896×19.09

= 74.375 m

The ball was in the air for 3.896 seconds

8 0

14 days ago

When 999mm is added to 100m ______ is the result

Keith_Richards [2877]

Answer:

The outcome of adding 999mm to 100m is 101m.

Explanation:

That's my belief.

6 0

1 month ago

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You are camping in the wilderness. After a few days, you are horrified to discover that you did not pack as many batteries as yo

Keith_Richards [2877]

Angular speed is calculated as 2.5 rev/sec.

8 0

13 days ago

A proton is released from rest at the origin in a uniform electric field that is directed in the positive xx direction with magn

Ostrovityanka [2807]

The alteration in potential energy is $\Delta PE = - 3.8*10^{-16} \ J$

In the query, it is stated that

The intensity of the uniform electric field equals $E = 950 \ N/C$

The distance the electron covers is $x = 2.50 \ m$

Typically, the force exerted on this electron is expressed mathematically as

$F = qE$

Where F signifies the force and q represents the charge of the electron, which is a fixed value of $q = 1.60*10^{-19} \ C$

Thus

$F = 950 * 1.60 **10^{-19}$

$F = 1.52 *10^{-16} \ N$

Generally, the work-energy theorem is mathematically framed as

$W = \Delta KE$

Where W denotes the work done on the electron by the electric field and $\Delta KE$ is the change in kinetic energy

Additionally, work done on the electron can also be described as

$W = F* x *cos( \theta )$

Where $\theta = 0 ^o$ assuming that the electron's movement aligns with the x-axis

So

$\Delta KE = F * x cos (0)$

Inserting values

$\Delta KE = 1.52 *10^{-16} * 2.50 cos (0)$

$\Delta KE = 3.8*10^{-16} J$

According to the conservation of energy

$\Delta PE = - \Delta KE$

Where $\Delta PE$ signifies the change in potential energy

Thus

$\Delta PE = - 3.8*10^{-16} \ J$

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10 days ago